# How do you find the power (1+sqrt3i)^4 and express the result in rectangular form?

Mar 6, 2017

${\left(1 + \sqrt{3} i\right)}^{4} = - 8 - 8 \sqrt{3} i$

#### Explanation:

One of the most interesting ways in complex numbers is that we can divide and multiply them easily, when we write them in polar form, as compared to the process of multiplication when complex numbers are written in rectangular form.

When numbers are written in rectangular form $z = a + b i$, we represent them on argand plane something like Cartesian plane, in polar form complex numbers are written in terms of $r$ and $\theta$ where $r$ is the length of the vector - better associated as absolute or modular value of $z$ and $\theta$ is the angle made with the real axis. This is akin to points marked as polar coordinates.

Thus the number $a + i b$ in polar form is written as $r \cos \theta + i r \sin \theta$, where $r = \sqrt{{a}^{2} + {b}^{2}}$ and $\theta = {\tan}^{- 1} \left(\frac{b}{a}\right)$.

For example, the number $1 + \sqrt{3} i$ can be written as $\left(2 , \frac{\pi}{3}\right)$ i.e. as $2 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)$. Note that $1 = 2 \cos \left(\frac{\pi}{3}\right)$ and $\sqrt{3} = 2 \sin \left(\frac{\pi}{3}\right)$

The beauty is that when complex numbers are written in polar form their product is a lot easier, as while you multiply all $r$'s, $\theta$'s are just added.

This leads to De Moivre's Theorem, which states that if $z = r \left(\cos \theta + i \sin \theta\right)$, ${z}^{n} = {r}^{n} \left(\cos n \theta + i \sin n \theta\right)$.

Hence as $1 + \sqrt{3} i = 2 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)$

${\left(1 + \sqrt{3} i\right)}^{4} = {2}^{4} \left(\cos \left(\frac{4 \pi}{3}\right) + i \sin \left(\frac{4 \pi}{3}\right)\right)$

= $16 \left(- \frac{1}{2} - \frac{\sqrt{3}}{2} i\right)$

= $- 8 - 8 \sqrt{3} i$